Invertible Integers under Multiplication/Corollary 1

Corollary to Invertible Integers under Multiplication

Let $a, b \in \Z$ such that $a b = 1$.

Then $a = b = \pm 1$.

Proof

Let $a b = 1$.

From Integer Multiplication Identity is One, $1$ is the identity element for $\Z$.

Then, by definition, both $a$ and $b$ are invertible elements of $\Z$ for multiplication.

So by Invertible Integers under Multiplication, $a = \pm 1$ and $b = \pm 1$.

As $-1 \times 1 = -1$, it follows that either:

$a = b = 1$

or

$a = b = -1$.

$\blacksquare$

Sources

• 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$